1164MediumArrayHash TableMathGreedy

The Fewest Moths in the Jar

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Trains the technique from

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

Some of the moths in a jar were asked how many other moths share their own pattern, and replies holds the answers given, one per moth asked. Moths that were not asked said nothing.

A moth answering x belongs to a group of exactly x + 1 moths sharing one pattern, and moths in different groups never share a pattern.

Return the fewest moths the jar could hold.

Examples

Example 1

Input: replies = [1, 1, 1]
Output: 4

Two of the three moths can share a group of two, but the third needs a group of two of its own, so a fourth moth must be there unasked.

Example 2

Input: replies = [0, 1, 2, 3]
Output: 10

Each answer names a different group size, so the four groups hold one, two, three and four moths.

Example 3

Input: replies = [2, 2, 2, 2]
Output: 6

Groups of three are named. Three of the four moths fill one group and the fourth opens another, which needs two more moths to fill it.

Constraints

1 <= replies.length <= 1000
0 <= replies[i] < 1000

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def num_rabbits(replies: list[int]) -> int:

Java

public int numRabbits(int[] replies)

The intended approach

Time O(n) over the replies.Space O(n) for the tally.

Two moths giving different answers can never share a pattern, since their groups are different sizes. So the answers can be handled one value at a time. Among the moths answering `x`, each belongs to a group of `x + 1`, and packing as many of them as possible into each group is what keeps the jar small. So the groups needed are the count of such moths divided by `x + 1`, rounded up, and every group holds a full `x + 1` moths whether or not all of them were asked. Adding that across the distinct answers gives the total. Rounding up is the whole point: leftover moths cannot be squeezed into a full group and must open a fresh one, which then has to be filled out with moths that were never asked.

Pythonentry point: num_rabbits

def num_rabbits(replies):
    tally = {}
    for value in replies:
        tally[value] = tally.get(value, 0) + 1
    total = 0
    for value in tally:
        size = value + 1
        groups = (tally[value] + size - 1) // size
        total += groups * size
    return total

The obvious version

Walk the replies in order keeping, for each answer given, how much room is left in the group opened most recently for it. A moth slots into that room when there is any, and otherwise opens a fresh group, adding its whole size to the jar. Costs one pass with a running room count per answer.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_num_rabbits(replies):
    total = 0
    room = {}
    for value in replies:
        left = room.get(value, 0)
        if left > 0:
            room[value] = left - 1
        else:
            total += value + 1
            room[value] = value
    return total

How this problem was checked

Reference solution executed against 15 test cases — 15 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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